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3 years ago

A line contains the points a(-9,y) and b(-7,-2) find the value of y so the line has a slope of -3

Mathematics

1 answer:

harina [27]3 years ago

3 0

Answer:

y = -8

Step-by-step explanation:

To find the value for y, use the slope formula:

$m = \frac{y_2-y_1}{x_2-x_1}$

Substitute m = 3 and (-9,y) and (-7,-2).

$-3 = \frac{y--2}{-7--9} \\\\-3 = \frac{y+2}{-7+9}\\\\ -3 = \frac{y+2}{2}\\\\-6 = y + 2\\\\-8 = y$

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4 years ago

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Natali5045456 [20]

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3 years ago

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The following table represents the annual sales of a bakery for the last 7 years since the grand opening

Effectus [21]

The table is a linear regression model, and the equation of the regression model is y = 0.24x + 0.77

<h3>The scatter plot that represents the table</h3>

See attachment for the required scatter plot

<h3>The best model of the scatter plot</h3>

From the attached scatter plot, we can see that the points are almost on a straight line

Hence, the best model that fits the scatter plot is a linear model

<h3>The equation of the regression model</h3>

Using a graphing calculator, we have the following calculation summary:

Sum of x = 28
Sum of y = 12.1
Mean X = 4
Mean Y = 1.7286
Sum of squares (SSX) = 28
Sum of products (SP) = 6.7
b = SP/SSX = 6.7/28 = 0.23929
a = MY - bMX = 1.73 - (0.24*4) = 0.77143

The regression equation is represented as:

y = bx + a

So, we have:

y = 0.23929x + 0.77143

Approximate

y = 0.24x + 0.77

Hence, the equation of the regression model is y = 0.24x + 0.77

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2 years ago

A 75-gallon tank is filled with brine (water nearly saturated with salt; used as a preservative) holding 11 pounds of salt in so

Debora [2.8K]

Let $A(t)$ = amount of salt (in pounds) in the tank at time $t$ (in minutes). Then $A(0) = 11$ .

Salt flows in at a rate

$\left(0.6\dfrac{\rm lb}{\rm gal}\right) \left(3\dfrac{\rm gal}{\rm min}\right) = \dfrac95 \dfrac{\rm lb}{\rm min}$

and flows out at a rate

$\left(\dfrac{A(t)\,\rm lb}{75\,\rm gal + \left(3\frac{\rm gal}{\rm min} - 3.25\frac{\rm gal}{\rm min}\right)t}\right) \left(3.25\dfrac{\rm gal}{\rm min}\right) = \dfrac{13A(t)}{300-t} \dfrac{\rm lb}{\rm min}$

where 4 quarts = 1 gallon so 13 quarts = 3.25 gallon.

Then the net rate of salt flow is given by the differential equation

$\dfrac{dA}{dt} = \dfrac95 - \dfrac{13A}{300-t}$

which I'll solve with the integrating factor method.

$\dfrac{dA}{dt} + \dfrac{13}{300-t} A = \dfrac95$

$-\dfrac1{(300-t)^{13}} \dfrac{dA}{dt} - \dfrac{13}{(300-t)^{14}} A = -\dfrac9{5(300-t)^{13}}$

$\dfrac d{dt} \left(-\dfrac1{(300-t)^{13}} A\right) = -\dfrac9{5(300-t)^{13}}$

Integrate both sides. By the fundamental theorem of calculus,

$\displaystyle -\dfrac1{(300-t)^{13}} A = -\dfrac1{(300-t)^{13}} A\bigg|_{t=0} - \frac95 \int_0^t \frac{du}{(300-u)^{13}}$

$\displaystyle -\dfrac1{(300-t)^{13}} A = -\dfrac{11}{300^{13}} - \frac95 \times \dfrac1{12} \left(\frac1{(300-t)^{12}} - \frac1{300^{12}}\right)$

$\displaystyle -\dfrac1{(300-t)^{13}} A = \dfrac{34}{300^{13}} - \frac3{20}\frac1{(300-t)^{12}}$

$\displaystyle A = \frac3{20} (300-t) - \dfrac{34}{300^{13}}(300-t)^{13}$

$\displaystyle A = 45 \left(1 - \frac t{300}\right) - 34 \left(1 - \frac t{300}\right)^{13}$

After 1 hour = 60 minutes, the tank will contain

$A(60) = 45 \left(1 - \dfrac {60}{300}\right) - 34 \left(1 - \dfrac {60}{300}\right)^{13} = 45\left(\dfrac45\right) - 34 \left(\dfrac45\right)^{13} \approx 34.131$

pounds of salt.

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2 years ago

A gift box measures 20 centimeters long, 8 centimeters wide and 3 centimeters deep. Which is the correct unit of measurement to

yKpoI14uk [10]

Meters cubed because it is a 3d object and you are finding the volume

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4 years ago

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